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sertanlavr [38]
3 years ago
8

Binary compounds are formed by ............... ............... elements.

Chemistry
1 answer:
goblinko [34]3 years ago
6 0

Answer: i think its A diatomic compound..

Explanation: hope i helped! sorry if im wrong!

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Please help me with this chemistry problem
Margarita [4]

Answer:

50 g of K₂CO₃ are needed

Explanation:

How many grams of K₂CO₃ are needed to make 500 g of a 10% m/m solution?

We analyse data:

500 g is the mass of the solution we want

10% m/m is a sort of concentration,  in this case means that 10 g of solute (K₂CO₃) are contained in 100 g of solution

Therefore we can solve this, by a rule of three:

In 100 g of solution we have 10 g of K₂CO₃

In 500 g of solution we may have, (500 . 10) / 100 = 50 g of K₂CO₃

6 0
3 years ago
g In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an
Marizza181 [45]

Answer:

See explanation

Explanation:

A titration involves the addition of a titrant to an analyte solution. It is a method of volumetric analysis.

When a particular volume of titrant is added, the colour changes to signal the end point of the reaction.

The point at which the colour changes is called the equivalence point. This is the point at which the amount of titrant added is just enough to completely neutralize the analyte solution.

Hence the volume NaOH that needs to be added to the beaker containing HCl to cause a colour change is the volume of NaOH that is just enough to completely neutralize the HCl solution.

4 0
3 years ago
What is the scientific notation of 68000
Kaylis [27]
68000 = 6.8 * 10000 = 6.8 * 10^4  

hope this helps? c;
3 0
3 years ago
If you mix 4.5 g of hcl in enough water to make 0.9 l (900 ml of solution, what is the molarity of the solution?
Elina [12.6K]
We first need to convert grams of hydrogen chloride to moles.

molarity = moles of solute/liters of solution

To convert from grams to moles, we need to divide by it's molecular mass.

4.5 g HCl * (1 mole HCl/ 36.46 g HCl) = 0.123 moles HCl

0.12 moles/0.9 L = 0.13 M HCl

If they gave you the volume as 900 mL, then that would be 3 sig figs. And the sig figs would be limited to 2 because the mass.If, however, they gave you the volume as 0.9 L, then that would be 1 sig fig. And the final answer would be limited to to 1 sig fig.
4 0
3 years ago
Phosphorous acid, h3po3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. calculate the ph f
Varvara68 [4.7K]

Answer:

Explanation:

(a)

Before the addition of KOH :-

Given pKa1 of H3PO3 = 1.30

we know , pKa1 = - log10Ka1

Ka1 = 10-pKa1

Ka1 = 10-1.30

Ka1 = 0.0501

similarly pKa2 = 6.70 ,therefore Ka2 = 1.99 x 10-7

because Ka1 >> Ka2 , therefore pH of diprotic acid i.e H3PO3 can be calculated from first dissociation only .

ICE table is :-

H3PO3 (aq) <-------------> H+ (aq) + H2PO3-(aq)

I 2.4 M 0 M 0 M

C - x + x + x

E (2.4 - x )M x M x M

x = degree of dissociation

Now expression of Ka1 is :

Ka1 = [ H+ ] [ H2PO3-] / [ H3PO3]

0.0501 = x2 / 2.4 - x

on solving for x by using quadratic formula , we have

x = 0.32

Now [ H+ ] = [ H2PO3-] = 0.32 M

pH = - log [H+]

pH = - log 0.32

pH = - ( - 0.495)

pH = 0.495

Hence pH before the addition of KOH = 0.495

(b)

After the addition of 25.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.025 L = 0.06 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

Now 0.06 moles of KOH is equal to the half of the moles required for the first equivalent point . therefore pH at this point is equal to pKa1 .

Hence pH = 1.30 M

(c)

After the addition of 50.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.050 L = 0.12 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

because Number of moles of H3PO4 = Number of moles of KOH

therefore , this point is the first equivalence point

and pH = pKa1 + pKa2 / 2

pH = 1.30 + 6.70 / 2

pH = 4.00

Hence pH = 4.00

(d)

After the addition of 75.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.075 L = 0.18 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

This is the half way of the second equivalence point , therefore pH is equal to pKa2 .

Hence pH = 6.70

5 0
3 years ago
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